## Wednesday, September 2, 2015

### Classic proof: $\dfrac{22}{7}\gt \pi$

Prove $\dfrac{22}{7}\gt \pi$.

There is a very classic and elegant proof for this inequality involving $\pi$ and its fraction representative $\dfrac{22}{7}$.

It's using the graphical and integration method to prove $\dfrac{22}{7}\gt \pi$.

Proof_that_22/7_> pi

It first recognizes the graph of $y=\dfrac{\left(x^4\left(1-x\right)^4\right)}{1+x^2}$ lies on the first quadrant on the interval $[0,\,1]$, as shown in the graph below:

Therefore, the definite integral [MATH]\int_{0}^{1} \dfrac{\left(x^4\left(1-x\right)^4\right)}{1+x^2}\,dx\gt 0[/MATH].

Now, our effort remains on evaluating the definite integral above:

[MATH]\int_{0}^{1} \dfrac{\left(x^4\left(1-x\right)^4\right)}{1+x^2}\,dx[/MATH]

[MATH]=\int_{0}^{1} \dfrac{x^8-4 x^7+6 x^6-4 x^5+x^4}{1+x^2}\,dx[/MATH]

[MATH]=\int_{0}^{1} \left(x^6-4 x^5+5 x^4-4 x^2+4-\dfrac{4}{1+x^2}\right)\,dx[/MATH]

[MATH]=\left[\dfrac{x^7}{7}-\dfrac{2x^6}{3}+x^5-\dfrac{4x^3}{3}+4x-4\arctan (x)\right]_0^1[/MATH]

[MATH]=\left[\dfrac{1}{7}-\dfrac{2}{3}+1-\dfrac{4}{3}+4-4\arctan (1)\right]-[\arctan 0][/MATH]

[MATH]=\left[\dfrac{22}{7}-4\dfrac{\pi}{4}\right]-[0][/MATH]

$=\dfrac{22}{7}-\pi$

Therefore, we have $\dfrac{22}{7}-\pi\gt 0$, which implies $\dfrac{22}{7}\gt \pi$ and the proof is then elegantly done.

Note that this classic proof and the beauty of it lies in finding the right integrand that must satisfy all the conditions (the area bounded by the function and $x=0$ and $x=1$ must be positive, and also, the result must yield a $\dfrac{22}{7}-\pi$).

There certainly have been many trials and errors before figuring out that beauty rational function that works like charm in proving this problem.

I want to say that is the spirit of each of us should acquire, we should toy with the idea by experimenting with the help of the top-notch technology in order to generate the rational function that works!

My hat is off to the person who created this very proof!